Question

15.
The mass of earth is 6 x 1024 kg. The distance
between the earth ans the sun is 1.5 x 1011 m.
If the gravitational force between the two is
3.5 x 1022 N, what is the mass of the sun ?
use G = 6.7 * 10-11 Nm²/kg?​

Answer 1

Correct Question:

The mass of earth is $\sf { 6 \times {10}^{24}}$ kg. The distance between the earth and the sun is $\sf { 1.5 \times {10}^{11}}$m. If the gravitational force between the two is $\sf {3.5 \times {10}^{22}}$ N, what is the mass of the sun ?

Use G = $\sf { \underline{6.67} \times {10}^{-11}}$ Nm²/kg.

Required Solution:

Given:

• Mass of the earth $\sf {m_1}$ = is $\sf { 6 \times {10}^{24}}$ kg

• Distance between the earth and sun (R) = $\sf { 1.5 \times {10}^{11}}$m

• Gravitational force between them (F) = $\sf {3.5 \times {10}^{22}}$ N

To find:

• Mass of the sun $\sf {m_2}$

Calculation:

We know that,

$\sf { \leadsto F = \dfrac{G{m}_{1}{m}_{2}}{{R}^{2}}}$

$\sf { \leadsto F \times {R}^{2} = G{m}_{1}{m}_{2}}$

$\sf { \leadsto \dfrac{F \times {R}^{2}}{G{m}_{1}}={m}_{2}}$

Now, insert the values in the above equation:

$\sf { \longrightarrow \dfrac{3.5 \times {10}^{22} \times {(1.5 \times {10}^{11})}^{2}}{ 6.67 \times {10}^{-11} \times 6 \times {10}^{24}}={m}_{2}}$

                       

$\sf { \longrightarrow \dfrac{3.5 \times {10}^{22} \times 1.5 \times {10}^{11} \times 1.5 \times {10}^{11} }{ 6.67 \times {10}^{-11} \times 6 \times {10}^{24}}={m}_{2}}$

                       

$\sf { \longrightarrow \dfrac{3.5 \times {10}^{22 + 11 + 11} \times 1.5 \times 1.5 }{ 6.67 \times {10}^{-11 + 24} \times 6 }={m}_{2}}$

                       

$\sf { \longrightarrow \dfrac{3.5 \times {10}^{44} \times 1.5 \times 1.5 }{ 6.67 \times {10}^{13} \times 6 }={m}_{2}}$

                       

$\sf { \longrightarrow \dfrac{3.5 \times {10}^{44 - 13} \times 1.5 \times 1.5 }{ 6.67 \times 6 }={m}_{2}}$

                       

$\sf { \longrightarrow \dfrac{3.5 \times {10}^{31} \times 1.5 \times 1.5 }{ 6.67 \times 6 }={m}_{2}}$

                       

$\sf { \longrightarrow \dfrac{35 \times {10}^{31} \times 15 \times 15 \times 100 }{ 667 \times 6 \times 10 \times 10 \times 10}={m}_{2}}$

                       

$\sf { \longrightarrow \dfrac{35 \times {10}^{31} \times 15 \times 15 \times {10}^{2} }{ 667 \times 6 \times {10}^{3} }={m}_{2}}$

                       

$\sf { \longrightarrow \dfrac{35 \times {10}^{33 - 3} \times 15 \times 15}{ 667 \times 6 }={m}_{2}}$

                       

$\sf { \longrightarrow \dfrac{35 \times {10}^{30} \times 15 \times 15}{ 667 \times 6 }={m}_{2}}$

                       

$\sf { \longrightarrow \dfrac{\cancel{7875} \times {10}^{30} }{ \cancel{4002} }={m}_{2}}$

                       

$\sf { \longrightarrow 1.96 \times {10}^{30} ={m}_{2}}$

Therefore, mass of the sun (m_2) is $\sf{1.96 \times {10}^{30} \: kg}$

__________________________

Answer 2

We have,

Mass of Earth = 6 × 10²⁴ kg

Distance between the Earth and Sun = 1 AU = 1.5 × 10¹¹ m

Gravitational force = 3.5 × 10²² N

Universal gravitational constant. = 6.7 × 10-¹² N-m²/kg²

We know,

F = GMm/d²

(Terms have their usual meanings)

⇒ m = Fd²/GM

⇒ m = [(3.5 × 10²² N)(1.5 × 10¹¹ m)²]/[(6.7 × 10-¹¹ Nm²/kg²)(6 × 10²⁴ kg)]

⇒ m = [3.5 × (1.5)²]/[(6.7 × 6)] × 10²²+²²+¹¹-²⁴ kg

⇒ m = 7.845/40.2 × 10³¹

⇒ m = 0.195 × 10³¹

⇒ m = 1.95 × 10³⁰ ≈ 2 × 10³⁰ kg {Answer}.